Question

Simplify: $\underset{n\to \infty }{lim}\left(\frac{1}{n}\{{\sec }^2\frac{\pi }{4n}+{\sec }^2\frac{2\pi }{4n}+\dots +{\sec }^2\frac{n\pi }{4n}\}\right)$

• A. $4/\pi$
• B. $2/\pi$
• C. $3/\pi$
• D. $5/\pi$

Answer 1

The correct option is A $4/\pi$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}{\sec }^2\frac{r.\pi }{4n}$

When $r=1,\frac{r}{n}=0$ and when, $r=n,\frac{r}{n}=1$

Hence, we get:-

${\int }_0^1{\sec }^2\frac{\pi x}{4}dx$

$=\frac{4}{\pi }{\left[\tan \frac{\pi x}{4}\right]}_0^1$

$=\frac{4}{\pi }$